The set of points with spherical coordinates of the form
\[(\rho, \theta, \phi) = \left( 1, \theta, \frac{\pi}{6} \right)\]forms a circle.  Find the radius of this circle.
If $P = \left( 1, \theta, \frac{\pi}{6} \right),$ and $P$ has rectangular coordinates $(x,y,z),$ then
\[\sqrt{x^2 + y^2} = \sqrt{\rho^2 \sin^2 \phi \cos^2 \theta + \rho^2 \sin^2 \phi \sin^2 \theta} = |\rho \sin \phi| = \frac{1}{2}.\]Hence, the radius of the circle is $\boxed{\frac{1}{2}}.$

[asy]
import three;

size(180);
currentprojection = perspective(6,3,2);

triple sphericaltorectangular (real rho, real theta, real phi) {
  return ((rho*Sin(phi)*Cos(theta),rho*Sin(phi)*Sin(theta),rho*Cos(phi)));
}

real t;
triple O, P;
path3 circ;

O = (0,0,0);
P = sphericaltorectangular(1,60,30);

circ = sphericaltorectangular(1,0,30);

for (t = 0; t <= 360; t = t + 5) {
  circ = circ--sphericaltorectangular(1,t,30);
}

draw(circ,red);
draw((0,0,0)--(1,0,0),Arrow3(6));
draw((0,0,0)--(0,1,0),Arrow3(6));
draw((0,0,0)--(0,0,1),Arrow3(6));
draw(surface(O--P--(P.x,P.y,0)--cycle),gray(0.7),nolight);
draw(O--P--(P.x,P.y,0)--cycle);
draw((0,0,0.5)..sphericaltorectangular(0.5,60,15)..sphericaltorectangular(0.5,60,30),Arrow3(6));
draw((0.4,0,0)..sphericaltorectangular(0.4,30,90)..sphericaltorectangular(0.4,60,90),Arrow3(6));

label("$x$", (1.1,0,0));
label("$y$", (0,1.1,0));
label("$z$", (0,0,1.1));
label("$\phi$", (0.2,0.2,0.6));
label("$\theta$", (0.6,0.3,0));
label("$P$", P, N);
[/asy]